Let $M$ be a non-zero $3\times 3$ matrix satisfying $M^3=0$, where $0$ is the $3\times 3$ zero matrix. Then



This is a multiple correct choice type question.More than one may be correct answers.

My attempt:
Let $M$ be a non-singular matrix,therefore its inverse exists.
Multiply both sides by $M^{-1}$ to get,$M^2=0$.
Again multiply both sides by $M^{-1}$ to get,$M=0$
But this is a contradiction.So $M$ is a singular matrix.Its inverse does not exist.
But i dont know how to solve further.Please help me.Thanks.

  • $\begingroup$ what exactly is O? Also, in your attempt, why are you assuming your matrix is non-singular? Is that part of the statement? $\endgroup$
    – user62748
    Oct 31, 2015 at 4:16
  • 2
    $\begingroup$ Note: it is not possible for $M$ to be non-singular, because then $det(M)\neq 0$, but also $0=det(0)=det(M^3)=det(M)^3$. $\endgroup$
    – vadim123
    Oct 31, 2015 at 4:23
  • $\begingroup$ Do you know about the Jordan normal form? $\endgroup$
    – Dylan
    Oct 31, 2015 at 4:28
  • $\begingroup$ BY $O$ do you mean $0$? $\endgroup$
    – 2'5 9'2
    Oct 31, 2015 at 5:42
  • $\begingroup$ By $O$,question means null matrix. $\endgroup$ Oct 31, 2015 at 7:55

3 Answers 3


All of above is really not necessary $$\left(\frac{M^2}{2}+M+I\right)\left(\frac{M^2}{2}-M+I\right)=I$$

It is easy from here onwards


$M^3=0$ implies $\det(M-xI)=-x^3$ (since $M$ satisfies its own characteristic equation). Decompose expression in question into linear factors: $$\det(\frac 1 2M^2\pm M+I)=\frac 1 8\det(M-x_{\pm1}I)\det(M-x_{\pm2}I)=\frac 1 8x_{\pm1}^3x_{\pm2}^3=\frac 1 82^3=1$$ where $x_{\pm}1,x_{\pm2}$ are roots of $x^2\pm2x+2=0$ and hence $x_{\pm1}x_{\pm2}=2$.


If we put $M$ in Jordan canonical form, for some invertible matrix $Q$ we will have $$M=QAQ^{-1}$$ where $$A=\left(\begin{smallmatrix}0&a&0\\0&0&b\\0&0&0\end{smallmatrix}\right)$$ where $a,b$ are each $0$ or $1$ (not both $0$ since $M\neq 0$). Hence, we write $$0.5M^2+M+I=0.5QAQ^{-1}QAQ^{-1}+QAQ^{-1}+QIQ^{-1}=$$ $$=Q(0.5A^2+A+I)Q^{-1}$$ Taking determinants, we get $$det(Q)det(0.5A^2+A+I)det(Q)^{-1}=det(0.5A^2+A+I)=1$$ since $$0.5A^2+A+I=\left(\begin{smallmatrix}1&a&\frac{ab}{2}\\0&1&b\\0&0&1\end{smallmatrix}\right)$$

The other three questions are similar.

  • $\begingroup$ Sir,cant this question be solved without Jordan normal form?Jordan form is not in our syllabus.@vadim123 $\endgroup$ Nov 20, 2015 at 14:03
  • 2
    $\begingroup$ It's difficult to know what is and is not on your syllabus when your question does not give context for the question. I gave one of several standard ways to solve this problem. $\endgroup$
    – vadim123
    Nov 20, 2015 at 15:01

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